Answer
$t = 1$ corresponds to neither a maximum nor a minimum.
Work Step by Step
Given $$ g(t)=\frac{e^{t}}{t^{2}+1} $$
Since
\begin{align*}
g'(x) &= \frac{\frac{d}{dt}\left(e^t\right)\left(t^2+1\right)-\frac{d}{dt}\left(t^2+1\right)e^t}{\left(t^2+1\right)^2}\\
&= \frac{e^t\left(t^2+1\right)-2te^t}{\left(t^2+1\right)^2}
\end{align*}
Then $f(x)$ has critical points when
\begin{align*}
g'(x)&=0\\
\frac{e^t\left(t^2+1\right)-2te^t}{\left(t^2+1\right)^2} &=0\\
e^t[\left(t^2+1\right)-2t]&=0\\
t^2-2t+1&=0
\end{align*}
Then $t=1$ is a critical point; to find the interval of increasing and decreasing regions, choose $t=0,\ t=2$
\begin{align*}
g'(0)&>0 \\
g'(2)& >0
\end{align*}
Hence, $g'(t) $ does not change sign and therefore $t = 1$ corresponds to neither a maximum nor a minimum.
